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Depression in Freezing Point

Question 2.32...

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Question 2.32

Calculate the depression in the freezing point of water when 10g of $C H_{3} C H_{2} C H C l C O O H$ is added to 250 g of water, $K_{a} = 1.4 \times 10^{- 3}$ ; $K_{f} = 1.86 K k g m o l^{- 1}$

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Solution

Step I: Calculation of degree of dissociation
Mass of solute = 10g
Molar mass of solute
$(C H_{3} C H_{2} C H C l C O O H) = (12 \times 4) + (1 \times 7) + (35.5) + (16 \times 2) = 48 + 7 + 35.5 + 32 = 122.5 g m o l^{- 1}$
Molal concentration of solution
$= \frac{\frac{M a s s}{M o l a r m a s s}}{M a s s o f s o l v e n t (k g)} = \frac{(10 g)}{(122.5 g m o l^{- 1}) \times (0.25 k g)} = 0.326 m$
If $α$ is the degree of dissociation of $C H_{3} C H_{2} C H C l C O O H$ , then
$\begin{matrix} C H_{3} C H_{2} C H C l C O O H ⇌ & C H_{3} C H C l C O O^{-} & + H^{+} I n i t i a l c o n c . C m o l^{- 1} k g & 0 & 0 A t e q u i l i b r i u m C (1 - α) & C α & C α \end{matrix}$
$∴ K_{a} = \frac{C α \times C α}{C (1 - α)} = C α^{2}$
[Considering $(1 - α) = 1$ or dilute solutions]
$α^{2} = \frac{K_{a}}{C}$ or $α = \sqrt{\frac{K_{a}}{C}}$
So, $α = \sqrt{\frac{1.4 \times 10^{- 3}}{0.326}} = \sqrt{42.9 \times 10^{- 4}} = 6.55 \times 10^{- 2} = 0.065$
Step II: Calculation of van't Hoff factor (f)
$\begin{matrix} C H_{3} C H_{2} C H C l C O O H ⇌ & C H_{3} C H_{2} C H C l C O O^{-} & + H^{+} I n i t i a l n o . o f m o l e s & 1 & 0 & 0 M o l e s a t e q u i l i b r i u m & 1 - α & α & α \end{matrix}$
Total number of moles after dissociation
$= (1 - α) + (α) + (α) = (1 + α)$
Van't Hoff fraction
(i) $= \frac{T o t a l n u m b e r o f m o l e s a f t e r d i s s o c i a t i o n}{N u m b e r o f m o l e s b e f o r e d i s s o c i a t i o n}$
$i = \frac{(1 + α)}{1} = (1 + α)$
$= 1 + 0.065 = 1.065$
Step III: Calculation of depression in freezing point $(Δ T_{f})$
$Δ T_{f} = i K_{f} m = (1.065) \times (1.86 K k g m o l^{- 1}) \times (0.326 m o l k g^{- 1})$
$Δ T_{f} = 0.65 K$

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QUESTION 2.33

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